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MOLES. A collection of objects. Mole:. A collection of Avogadro’s number of objects. Avogadro’s number. 6.022 * 10 23. Atoms Particles Ions Cations Anions Molecules Formula Units. Mole Ratio.

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Moles
MOLES

A collection of objects

Mole:

A collection of Avogadro’s number of objects

Avogadro’s number

6.022 * 1023

Atoms

Particles

Ions

Cations

Anions

Molecules

Formula

Units


Mole Ratio

The coefficient in front of each component in a balanced equation

2 Al (s) + 6 HCl (aq) ----- 2 AlCl 3(aq) + 3 H2(g)

2 moles of Al reacts w/ 6 moles HCl

to produce

2 moles AlCl3 and 3 moles H2 gas

This states:


1 mole

Argon

atoms

1 mole

Helium

atoms

MOLE CALCULATIONS

One mole of argon atoms ….

….. is a number

One mole of helium atoms ….

….. is a number

6.022 * 1023 Argon atoms

6.022 * 1023 Helium atoms

….. has a mass

….. has a mass

39.9 grams

4.00 grams


RECAP: 1 mol of “anything” contains 6.022*1023 “parts”

Elements on p.table = I mol

Molar mass (mass/1 mol)

1. Elements

- decimal # on p.table

Ti 47.87 amu or g

2. Molecules/Cmpd

- sum of mass of all elements


MOLES

mass = 1 mole

Divide

by

Multiply

by

MASS


Given mass, find moles

How many moles are in 345.6 g NaNO3 ?

Step 1: find formula wt. of NaNO3

1 Na = 23.0 g

1 N = 14.0 g

3 O = 3 * 16.0 = 48.0 g

1 mole of NaNO3 = 85.0 g

Step 2 : Find # of moles

Use factor-label method

4.07 moles NaNO3


Given moles, find mass

How many grams are in 0.6 moles N2O ?

Step 1: find molecular wt. of N2O

2 N = 2 * 14.0 = 28.0 g

1 O = 16.0 g

1 mole of N2O = 44.0 g

Step 2 : Find mass

Use factor-label method

26.4 g N2O


Multiply

by

Divide

by

MOLES

6.02*1023

Avogadro’s Number

NUMBER of

PARTICLES


Given moles, find # molecules

How many molecules are in 1.6 moles oxygen?

Step 1: Recognize that mass does not apply here.

But Avogadro’s # is used

Step 2 : Find # molecules

Use factor-label method

9.63 * 1023 molecules O2


Given # atoms, find moles

How many moles are in 3.01 *1012 atoms of Chromium?

Step 1: Recognize that mass does not apply here.

But Avogadro’s # is used

Step 2 : Find # moles

Use factor-label method

5.00 * 10-12 mols Cr


Multiply

by

Divide

by

MOLES

6.02*1023

mass = 1 mole

Avogadro’s Number

Divide

by

Multiply

by

MASS

NUMBER of

PARTICLES

There is not a direct relationship

between MASS and PARTICLES


2 Step Conversion Problem

Will need to use MASS at some

point in the problem

Given # formula units, find grams

How many grams are in 1.208 * 1024 formula units AgCl ?

Step 1: find formula wt. of AgCl

1 Ag = 107.9 g

1 Cl = 35.5 g

1 mole of AgCl = 143.4 g

Step 2 : Find moles

Step 3 : Find mass

Use factor-label method

288 g AgCl

Converts moles

to MASS

Converts formula units

to MOLES


2 Step Conversion Problem

Will need to use MASS at some

point in the problem

Given grams, find # atoms

How many atoms are in 128.0 grams of Mercury ?

Step 1: find molar mass of Hg

1 mole of Hg = 200.6 g

Step 2 : Find moles

Step 3 : Find atoms

Use factor-label method

3.84*1023 atoms Hg

Converts mass

to MOLES

Converts moles

to ATOMS


+

PROBLEM

What is the formula weight of sodium carbonate,

Na2CO3. This is an industrial chemical used in

making glass.

Also, equivalent to 1mole

of a substance

SOLUTION

2 * 23.0 = 46.0

2 Na

1 C

3 O

1 * 12.0 = 12.0

3 * 16.0 = 48.0

106.0 g


BALANCE EQUATIONS

“STOICHIOMETRY”

Sodium Phosphate (aq) + Barium Nitrate (aq) ------

Barium Phosphate (s) + Sodium Nitrate (aq)

2

Na3PO4(aq) + Ba(NO3)2(aq) ----- Ba3(PO4)2(s) + NaNO3 (aq)

3

3

6

Aluminum (s) + Hydrochloric Acid (aq) ---------->

Aluminum Chloride (aq) + Hydrogen (g)

3

Al + HCl ---- AlCl3 + H2

2

6

2

3


MASS - MASS CALCULATIONS & LIMITING REAGENT

(amts of 2 reactants given: limiting reactants)

Solid calcium metal burns in air (oxygen) to form a calcium oxide cmpd.

Using 4.20 g Ca & 2.80 g O2 how much pdt is made?

Find 1) grams of pdt made from g of Ca

2) grams of pdt made from g of O2

3) limiting reagent

4) how much pdt can be made

5) % yield if 3.85 g produced

PLAN: Need ------> balanced chem. rxn

STEPS:

g Ca -----> mols Ca -----> mols pdt ------> g pdt

M g Ca coeff X pdt/Y react M g pdt

(same steps converting oxygen)

2 2

Ca(s) + O2(g) -------> CaO(s)

M g Ca M g O2 coeff X pdt/Y react M g pdt

40.1 g Ca:CaO 2:2 56.1 g

32.0 g O2:CaO 1:2


4.20 g 2.80 g X g

5.89 g

9.82 g

2 Ca(s) + O2(g) -------> 2 CaO(s)

M g : 40.1g 56.1 g

Coeff : 2 1 2

mols :

0.105

0.175

0.105

0.0875

3) limiting reagent:

Which produced the least amt-- Ca or O2?

Ca ---> CaO = 5.89 g O2 ---> CaO = 9.82 g

Ca, limiting reagent

5.89 g CaO

4) how much pdt can be made:

5) % yield=


RECAP: -- Balanced chem. equation --

then,GIVENUSEFIND

grams Molar Mass mols

mols coeff mols

mols Molar Mass grams

limiting reagent: the reactant that produces the least

amt of moles or mass of a specific pdt


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